Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 7;
vector<int> v[N];
vector<int> x[20][N];
vector<vector<int> > res;
int dep[N];
int par[20][N];
int n, m, q, foo, bar, k;
void dfs(int node, int baap) {
  par[0][node] = baap;
  dep[node] = dep[baap] + 1;
  for (int i = 0; i < v[node].size(); i++) {
    int beta = v[node][i];
    if (beta == baap) continue;
    dfs(beta, node);
  }
}
int jump(int p, int dist) {
  for (int i = 19; i >= 0; i--) {
    if ((1 << i) <= dist) {
      dist -= (1 << i);
      p = par[i][p];
    }
  }
  return p;
}
int lca(int p, int q) {
  if (dep[p] < dep[q]) swap(p, q);
  p = jump(p, dep[p] - dep[q]);
  if (p == q) return p;
  for (int i = 19; i >= 0; i--) {
    int pp = par[i][p];
    int qq = par[i][q];
    if (pp != qq) p = pp, q = qq;
  }
  return par[0][p];
}
vector<int> merge(vector<int> &a, vector<int> &b, int lim) {
  vector<int> temp;
  int i, j;
  for (i = 0, j = 0; i < a.size() and j < b.size() and temp.size() < lim;) {
    if (a[i] == b[j])
      temp.push_back(a[i]), i++, j++;
    else if (a[i] > b[j])
      temp.push_back(b[j++]);
    else
      temp.push_back(a[i++]);
  }
  while (i < a.size() and temp.size() < lim) temp.push_back(a[i++]);
  while (j < b.size() and temp.size() < lim) temp.push_back(b[j++]);
  return temp;
}
void update(int node, int dist, int k) {
  for (int i = 19; i >= 0; i--) {
    if ((1 << i) <= dist) {
      dist -= (1 << i);
      res.push_back(x[i][node]);
      node = par[i][node];
    }
  }
}
int main() {
  cin >> n >> m >> q;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &foo, &bar);
    v[foo].push_back(bar);
    v[bar].push_back(foo);
  }
  for (int i = 0; i < m; i++) {
    scanf("%d", &foo);
    x[0][foo].push_back(i + 1);
  }
  for (int i = 1; i <= n; i++) {
    sort(x[0][i].begin(), x[0][i].end());
    if (x[0][i].size() > 10) x[0][i].resize(10);
  }
  dfs(1, 0);
  for (int i = 0; i < 19; i++)
    for (int j = 1; j <= n; j++) {
      par[i + 1][j] = par[i][par[i][j]];
      x[i + 1][j] = merge(x[i][j], x[i][par[i][j]], 10);
    }
  while (q--) {
    scanf("%d%d%d", &foo, &bar, &k);
    int l = lca(foo, bar);
    res.clear();
    update(foo, dep[foo] - dep[l] + 1, k);
    update(bar, dep[bar] - dep[l], k);
    set<pair<int, int> > s;
    vector<int> ans;
    vector<int> ind(res.size());
    for (int i = 0; i < res.size(); i++)
      if (res[i].size()) s.insert(make_pair(res[i][0], i));
    while (!s.empty() and ans.size() < k) {
      int xx = s.begin()->first;
      int yy = s.begin()->second;
      s.erase(s.begin());
      ans.push_back(xx);
      ind[yy]++;
      if (ind[yy] < res[yy].size()) s.insert(make_pair(res[yy][ind[yy]], yy));
    }
    k = ans.size();
    printf("%d ", k);
    for (int i = 0; i < k; i++) printf("%d ", ans[i]);
    puts("");
  }
  return 0;
}